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Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane $\left( C _3 H _8\right)$ is burned? Express your answer to the correct number of significant figures and show all work in the form of dimensional analysis.



$C_3 H_8+5 O_2 \longrightarrow 3 CO_2+4 H_2 O$

Verified Answer :

  • Determine molar masses for propane (\(44.094 \text{ g/mol}\)) and carbon dioxide (\(44.01 \text{ g/mol}\)).
  • Set up a dimensional analysis expression: \(100.00 \text{ g } \text{C}_3\text{H}_8 \times \frac{1 \text{ mol } \text{C}_3\text{H}_8}{44.094 \text{ g } \text{C}_3\text{H}_8} \times \frac{3 \text{ mol } \text{CO}_2}{1 \text{ mol } \text{C}_3\text{H}_8} \times \frac{44.01 \text{ g } \text{CO}_2}{1 \text{ mol } \text{CO}_2}\).
  • Calculate the result: \(\approx 299.42849 \text{ g}\).
  • Round to five significant figures to get \(\boxed{299.43 \text{ g}}\).

Explanation

  1. Problem Analysis

To determine the mass of carbon dioxide (\(\text{CO}_2\)) produced by the combustion of \(100.00 \text{ g}\) of propane (\(\text{C}_3\text{H}_8\)), we use the balanced chemical equation: \[\text{C}_3\text{H}_8 + 5\text{O}_2 \longrightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\] This tells us that \(1 \text{ mole}\) of propane produces \(3 \text{ moles}\) of carbon dioxide. We are given the initial mass of propane as \(100.00 \text{ g}\), and we use the standard atomic masses of carbon (\(12.01 \text{ g/mol}\)), hydrogen (\(1.008 \text{ g/mol}\)), and oxygen (\(16.00 \text{ g/mol}\)) to determine the molar masses required for our conversion calculations.

  1. Calculating Molar Masses

First, we calculate the molar masses of the substances involved:

  • The molar mass of propane (\(\text{C}_3\text{H}_8\)) is:

\[3 \times (12.01 \text{ g/mol}) + 8 \times (1.008 \text{ g/mol}) = 36.03 \text{ g/mol} + 8.064 \text{ g/mol} = 44.094 \text{ g/mol}\]

  • The molar mass of carbon dioxide (\(\text{CO}_2\)) is:

\[1 \times (12.01 \text{ g/mol}) + 2 \times (16.00 \text{ g/mol}) = 12.01 \text{ g/mol} + 32.00 \text{ g/mol} = 44.01 \text{ g/mol}\]

  1. Dimensional Analysis and Final Result

Next, we perform the dimensional analysis to convert grams of propane to grams of carbon dioxide:
\[100.00 \text{ g } \text{C}_3\text{H}_8 \times \left(\frac{1 \text{ mol } \text{C}_3\text{H}_8}{44.094 \text{ g } \text{C}_3\text{H}_8}\right) \times \left(\frac{3 \text{ mol } \text{CO}_2}{1 \text{ mol } \text{C}_3\text{H}_8}\right) \times \left(\frac{44.01 \text{ g } \text{CO}_2}{1 \text{ mol } \text{CO}_2}\right)\]
Multiplying these factors:
\[\frac{100.00 \times 1 \times 3 \times 44.01}{44.094 \times 1 \times 1} \text{ g } \text{CO}_2 = \frac{13203}{44.094} \text{ g } \text{CO}_2 \approx 299.42849 \text{ g } \text{CO}_2\]
Rounding this result to five significant figures gives \(\boxed{299.43 \text{ g}}\).

Examples

This method is essential in environmental science for calculating emissions from fuel combustion. For example, by knowing how much fuel is burned in a power plant or a vehicle, researchers can determine the exact mass of CO2, a greenhouse gas, that is released into the atmosphere, which is crucial for tracking carbon footprints and climate impact.

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